Question: Simplify the following expression and state the condition under which the simplification is valid. $k = \dfrac{-2x^3 + 6x^2 + 36x}{-10x^3 + 70x^2 - 60x}$
First factor out the greatest common factors in the numerator and in the denominator. $ k = \dfrac {-2x(x^2 - 3x - 18)} {-10x(x^2 - 7x + 6)} $ $ k = \dfrac{2x}{10x} \cdot \dfrac{x^2 - 3x - 18}{x^2 - 7x + 6} $ Simplify: $ k = \dfrac{1}{5} \cdot \dfrac{x^2 - 3x - 18}{x^2 - 7x + 6}$ Since we are dividing by $x$ , we must remember that $x \neq 0$ Next factor the numerator and denominator. $ k = \dfrac{1}{5} \cdot \dfrac{(x - 6)(x + 3)}{(x - 6)(x - 1)}$ Assuming $x \neq 6$ , we can cancel the $x - 6$ $ k = \dfrac{1}{5} \cdot \dfrac{x + 3}{x - 1}$ Therefore: $ k = \dfrac{ x + 3 }{ 5(x - 1)}$, $x \neq 6$, $x \neq 0$